A 10V battery with internal resistance 1 Ω and a 15V battery with internal resistance 0.6Ω are connected in parallel to a voltmeter. The reading in the voltmeter will be close to.
A
11.9 V
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B
13.1 V
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C
24.5 V
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D
12.5 V
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Solution
The correct option is B 13.1 V Applying KVL in the circuit 15−10−1∗i−.6∗i=0 The current through voltmeter will be negligible as voltmeter have very high resistance. ⇒i=51.6=258 The voltmeter reading is given as potential difference across points A and B ΔV=−i∗.6+15=15−258∗.6=15−1.875=13.12V reading across voltmeter will be 13.2 V, correct option is B.